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Re: st: Average of Sample vesus Sub-Samples


From   Tim Scharks <tim.scharks@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Average of Sample vesus Sub-Samples
Date   Fri, 19 Mar 2010 09:05:49 -0700

Ah, I see That's what I get for faulty inference and reading too fast...thanks!

On Fri, Mar 19, 2010 at 8:59 AM, Jeph Herrin <junk@spandrel.net> wrote:
> The original question was not whether these two
> numbers could differ, but whether the average of the
> larger group could be greater than the mean of the
> two sub averages. In your case they differ, but the
> average of the larger group is smaller.
>
> I think you have to have negative numbers to reverse
> the inequality.
>
> cheers,
> Jeph
>
>
> Tim Scharks wrote:
>>
>> but what about
>>
>> A = 1, 1, 1
>> B = 3
>> m(A) = 1
>> m(B) = 3
>> (m(A)+m(B))/2=2
>> mean (A+B) =1.5
>>
>> uh oh...
>>
>> The problem is not negative numbers--it is your failure to weight the
>> subsample means according to their relative size:
>>
>> (m(A)*3+m(B)*1)/4 =1.5
>> m(A+B) = 1.5
>>
>>
>> On Fri, Mar 19, 2010 at 6:49 AM, Jeph Herrin <junk@spandrel.net> wrote:
>>>
>>> Yes, if they are negative.
>>>
>>>
>>>  A = -1,-1,-1
>>>  B = -3
>>> then
>>>  mean(A) = -1
>>>  mean(B) = -3
>>>  (mean(A)+mean(B))/2 = -2
>>>  mean(A+B)= -6/4 = -1.5
>>>
>>> -1.5 > -2
>>>
>>> So the average of all 4 numbers is larger than
>>> the average of the means of the two subsamples.
>>>
>>>
>>> hth,
>>> Jeph
>>>
>>>
>>>
>>>
>>>
>>> Erasmo Giambona wrote:
>>>>
>>>> Dear Statalisters,
>>>>
>>>> Is it possible that the arithmetic average of a sample is larger than
>>>> the averages of two sub-samples containing overall all the
>>>> observations of the full samples?
>>>>
>>>> Any thoughts would be appreciated,
>>>>
>>>> Erasmo
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